Subject: [mg62771] Re: How would you evaluate this limit in Mathematica

From: danl at wolfram.com

Date: Mon, 5 Dec 2005 03:37:12 -0500 (EST)

References: <dmpa43$hl7$1@smc.vnet.net><dmtc4k$fpj$1@smc.vnet.net>

Sender: owner-wri-mathgroup at wolfram.com

Robert Knighten wrote:
> Robert Knighten <Robert at Knighten.org> writes:
>
> > There has recently been an off-topic discussion on the category theory
> > mailing list of the assertion that computer algebra systems are tricky to
> > use correctly. The starting point was the question of evaluating the
> > (parametric) limit:
> >
> > Lim ((1 + 4 x^2)^(1/4) - (1 + 5 x^2)^(1/5))
> > x->0 ---------------------------------------
> > (a^(-x^2/2) - Cos[x])
> >
> > So I am interested in getting comments about doing this in Mathematica. I'm
> > also interested in how one would teach solving limits using Mathematica.
>
> My thanks for the various replies.
>
> The reason this is an interesting question is because the limit is indeed 0
> EXCEPT for the one exceptional case when a = E where the limit is 6. The
> problem was originally given to students in a mathematical analysis course as
> part of the early discussion of limits, and was then taken by one of the
> students as a test case to another class where they were being introduced to
> CAS. [I'm just an interested bystander in all of this.]
>
> So Mathematica gives 0 for
>
> Limit[
> ((1 + 4 x^2)^(1/4) - (1 + 5 x^2)^(1/5))/
> (a^(-x^2/2) - Cos[x]), x->0]
>
> but does indeed give 6 for
>
> Limit[
> ((1 + 4 x^2)^(1/4) - (1 + 5 x^2)^(1/5))/
> (E^(-x^2/2) - Cos[x]), x->0]
>
> but how is a Mathematica user supposed to learn that?
>
> This limit is a 0/0 case, so one can use L'Hospital's rule
>
> Limit[D[((1 + 4 x^2)^(1/4) - (1 + 5 x^2)^(1/5)),x]/D[(a^(-x^2/2) -
> Cos[x]),x],x->0]
>
> which again gives 0. A good student will remember to check and will find that
>
> Limit[D[((1 + 4 x^2)^(1/4) - (1 + 5 x^2)^(1/5)),x], x->0] is 0
>
> and
>
> Limit[D[(a^(-x^2/2) - Cos[x]),x],x->0], x->0] is 0 as well.
>
> Iterating L'Hospital's Rule doesn't help -- the answer is still 0, but the
> limit of the second derivative of the denominator is 1 - Log[a] which should
> give a hint that the special case where 1 - Log[a] = 0 needs to be
> investigated.
>
>
> An easier method is to look at
>
> Series[((1 + 4 x^2)^(1/4) - (1 + 5 x^2)^(1/5))/
> (a^(-x^2/2) - Cos[x]), {x, 0, 3}]
>
> which gives
>
> x^2/(1 - Log[a]) + O[x]^4
>
> and shows there is a problem when Log[a] = 1, i.e. when a = E, and then
>
> Series[((1 + 4 x^2)^(1/4) - (1 + 5 x^2)^(1/5))/
> (E^(-x^2/2) - Cos[x]), {x, 0, 3}]
>
> gives
>
> 6 - 143x^2/5 + O[x]^4
>
> which shows the answer in that special case, but this is certainly what I
> *would* call tricky. And this is long before a discussion of Taylor series.
>
>
>
> -- Bob
>
> --
> Robert L. Knighten
> Robert at Knighten.org
Probably all the more time before discussion of points of indeterminacy
(most students will not encounter that concept unless and until taking
a course in several complex variables).
You have a meromorphic function in {a,x} in a neighborhood of {E,0}. At
that particular point the limiting value will be path dependent. For
example,
func[a_, x_] := ((1 + 4x^2)^(1/4) - (1 + 5* x^2)^(1/5)) /(a^(-x^2/2) -
Cos[x])
In[26]:=Limit[func[E+m*t^2,t],t->0]
Out[27]=(6*E)/(E - 6*m)
We can thus attain any nonzero limit by judicious selection of m, if we
have the second variable approach 0 linearly and the first approach E
quadratically.
I don't particularly disagree with sentiments expressed in several
responses, regarding how to use or not use symbolic computation in
education. But if one point stands out from an example such as this
one, it is that there can be subtleties both to the math and the
interaction with symbolic algebra. I do not think this necessarily
indicates a deficiency in how we teach, or how we use symbolic programs
(in and out of the classroom), or in how such programs work. Which is
not to say that other reasons do not exist to indicate flaws in these
various areas.
Daniel Lichtblau
Wolfram Research